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Rank and Nullity are essential concepts in linear algebra, particularly in the context of matrices and linear transformations. They help describe the number of linearly independent vectors and the dimension of the kernel of a linear mapping.
For the following matrix A of order 5 Γ 5, the rank of A is 4, and the nullity of A is 1.
Nullspace of any matrix is defined as the solution associated with the system of homogenous equation AX = O where A is any real matrix of order, m Γ n.
Nullspace of A = { x β Rn | Ax = O}. Then the nullity of A is the dimension of the Nullspace of A.
The rank and nullity of a matrix can be calculated using the following steps:
Let T: V β W be a linear transformation, where V is a finite-dimensional vector space. Then,
rank(T) + nullity(T) = dim(V)
Equivalently, dim(V) = dim(Ker(T)) + dim(Im(T))
For a matrix A with n columns, Rank(A) + Nullity(A) = n
where Rank(A) is the dimension of the image (column space) and Nullity(A) is the dimension of the null space (kernel).
Let N = Ker(T). Since N is a subspace of V and V is finite-dimensional, N is also finite-dimensional.
Let dim(N) = k and let {Ξ±β, Ξ±β, ..., Ξ±β} be a basis of N.
Since these vectors are linearly independent in V, they can be extended to form a basis of V: {Ξ±β, Ξ±β, ..., Ξ±β, Ξ±βββ, ..., Ξ±β}
where dim(V) = n.
Now consider the vectors: T(Ξ±βββ), T(Ξ±βββ), ..., T(Ξ±β)
We will show that these vectors form a basis for Im(T).
Step 1: They span Im(T).
Let Ξ² be any vector in Im(T). Then there exists a vector Ξ± in V such that:
T(Ξ±) = Ξ²
Since the Ξ±'s form a basis of V,
Ξ± = aβΞ±β + aβΞ±β + ... + aβΞ±β
Applying T,
Ξ² = aβT(Ξ±β) + aβT(Ξ±β) + ... + aβT(Ξ±β)
Because Ξ±β, Ξ±β, ..., Ξ±β belong to Ker(T),
T(Ξ±β) = T(Ξ±β) = ... = T(Ξ±β) = 0
Therefore, Ξ² = aβββT(Ξ±βββ) + ... + aβT(Ξ±β)
Hence, T(Ξ±βββ), ..., T(Ξ±β) span Im(T).
Step 2: They are linearly independent.
Suppose, cβββT(Ξ±βββ) + ... + cβT(Ξ±β) = 0
Using linearity of T,
T(cβββΞ±βββ + ... + cβΞ±β) = 0
So, cβββΞ±βββ + ... + cβΞ±β β Ker(T)
Since every vector in Ker(T) can be written as a linear combination of Ξ±β, Ξ±β, ..., Ξ±β,
cβββΞ±βββ + ... + cβΞ±β = bβΞ±β + ... + bβΞ±β
Rearranging, bβΞ±β + ... + bβΞ±β β cβββΞ±βββ β ... β cβΞ±β = 0
Since {Ξ±β, Ξ±β, ..., Ξ±β} is a basis of V, these vectors are linearly independent. Therefore,
bβ = bβ = ... = bβ = cβββ = ... = cβ = 0
Hence, T(Ξ±βββ), ..., T(Ξ±β) are linearly independent.
Therefore, T(Ξ±βββ), ..., T(Ξ±β) form a basis of Im(T).
So, Rank(T) = dim(Im(T)) = n β k and Nullity(T) = k.
Adding them,
Rank(T) + Nullity(T) = (n β k) + k = n
Since n = dim(V),
Rank(T) + Nullity(T) = dim(V)
Hence proved.
The rank and nullity of a matrix have various applications in linear algebra, including:
Some examples on rank and nullity are,
Example 1: Given Matrix
Find the rank and nullity of B.
Solution:
Using Row Transformation in matrix B,
R2 β R3 - 2R2
Now, R3 β R3 - 4R1
Now, R3 β 3R2 + R3
β΄ r (B) = 2.
n (B) = n (columns) - r (B) = 4 - 2 = 2.
β΄ Rank of matrix B is 2 and the nullity of matrix B is 2.
Example 2: Given Matrix
Find the rank of matrix A.
Solution:
Using Row Transformation in matrix A,
R3 β R3 + R1
Now, R2 β R2 - 3R1
Now, R3 β R3 + R2
r (A) = 2
β΄ Rank of matrix A is 2.
Example 3: Given Matrix
Find the nullity of matrix D.
Solution:
Using Row Transformation in matrix D,
R3 β R3 - 5R1
Now, R4 β 2R1 + R4
Now, R3 β -8R2 + R3
Now, R4 β 9R2 + 2R4
Now, R2 β -1/2 R2
r (D) = 2
n (D) = n (columns) - r (D) = 2 - 2 = 0.
β΄ Nullity of matrix D is 0.